Question

If $z=\frac{\sqrt{3}+i}{\sqrt{3}-i}$ then the fundamental amplitude of z is

• A. $-\frac{\pi }{3}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{3}$

Given, $z=\frac{\sqrt{3}+i}{\sqrt{3}-i}$

$=\frac{\sqrt{3}+i}{\sqrt{3}-i}\cdot \frac{\sqrt{3}+i}{\sqrt{3}+i}$

$=\frac{3+2\sqrt{3}i+i^2}{3-i^2}$
$\Rightarrow z=\frac{2+2\sqrt{3}i}{4},\because i^2=-1$
$\Rightarrow z=\frac{1+\sqrt{3}i}{2}$
$\therefore \text{amp(z)}={\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$